Question

A particle of mass m = 8 x 10$^{-26}$ kg and charge q = 16 × 10$^{-19}$ C enters in a region of uniform magnetic field of strength 0.5 T at point A and leaves at point B with speed 10$^6$ m/s as shown in figure. The value of distance AB is

Answer 1

0.1 m

Answer 2

To solve this problem, we need to determine the radius of the circular path of the charged particle in the magnetic field and then use the geometry of the situation to find the distance AB.

1. Calculate the radius of the circular path (r):

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force provides the necessary centripetal force.

The magnetic force is given by F_B = qvB sin(θ).

The centripetal force is given by F_c = mv^2/r.

In this case, the velocity of the particle is perpendicular to the magnetic field (velocity is in the plane of the paper, magnetic field is into the page), so θ = 90° and sin(90°) = 1.

Therefore, qvB = mv^2/r.

From this, the radius r can be calculated as:

r = mv / (qB)

Given values:

• Mass, m = 8 × 10^-26 kg
• Charge, q = 16 × 10^-19 C
• Magnetic field strength, B = 0.5 T
• Speed, v = 10^6 m/s

Substitute these values into the formula for r:

r = (8 × 10^-26 kg × 10^6 m/s) / (16 × 10^-19 C × 0.5 T)

r = (8 × 10^(-26 + 6)) / (16 × 0.5 × 10^-19)

r = (8 × 10^-20) / (8 × 10^-19)

r = 1 × 10^(-20 - (-19))

r = 1 × 10^-1 m

r = 0.1 m

2. Analyze the geometry to find distance AB:

The particle enters the magnetic field at point A and leaves at point B, both of which lie on the vertical dashed line that forms the boundary of the magnetic field.

The initial velocity vector at point A makes an angle of 60° with this vertical boundary line.

Let α be the angle of incidence with the normal to the boundary. The normal to the vertical boundary is a horizontal line.

So, α = 90° - 60° = 30°.

The path of the particle inside the uniform magnetic field is a circular arc. Let C be the center of this circular path.

The lines connecting the center C to the entry point A (CA) and the exit point B (CB) are radii of the circle, so CA = CB = r.

The angle subtended by the chord AB at the center of the circle is θ = 2α.

θ = 2 × 30° = 60°.

Now consider the triangle ΔACB. It is an isosceles triangle with CA = CB = r and the angle between these two sides ∠ACB = 60°.

An isosceles triangle with the angle between its equal sides being 60° is an equilateral triangle.

Therefore, all sides of ΔACB are equal: AB = CA = CB = r.

3. Calculate the distance AB:

Since AB = r, and we calculated r = 0.1 m.

AB = 0.1 m

Note on consistency: The problem states q = 16 × 10^-19 C (a positive charge) and the magnetic field is into the page (indicated by 'x'). If the velocity is initially up-right (as shown by the arrow), a positive charge would experience a force directed left and upwards (using Fleming's Left Hand Rule or F = q(v x B)). This would cause the particle to curve upwards and to the left. However, the diagram shows point B below point A, implying a downward curvature. This indicates an inconsistency between the stated positive charge/field direction and the depicted trajectory. Nevertheless, for such problems, we proceed assuming the geometric configuration (entry/exit points and angles) is accurate for determining the distance AB. The radius calculation and the geometric relation AB = r are independent of the direction of curvature.